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When the velocity depends on $u$, different initial values travel at different speeds. The resulting characteristic crossing can destroy a classical solution in finite time, even though the initial function is smooth and bounded. For the Inviscid Burgers equation, take $u_0(a)=-\tanh a$. Before crossing,andAt $a=0$ this tends to $-\infty$ as $t\uparrow 1$, while $|u|\le 1$. Thus $\boxed{\text{bounded states can undergo finite-time gradient blow-up and shock formation}.}$ The bounded-derivative Burgers flux agrees with this equation on the whole state range, if a global bounded-speed flux is desired. After crossing, arbitrary weak solutions need not be unique; an entropy solution condition is needed to select the admissible continuation.

A bounded weak solution satisfiesfor every compactly supported $C^1$ test function, of either sign, allowed to meet $t=0$. Across a moving jump $x=s(t)$ with distinct left/right constant states, the distributional delta coefficient vanishes exactly when the Rankine-Hugoniot condition holds:It is required on every interface of a piecewise constant solution; the initial trace must also agree with the data.

For nonuniqueness respecting the paper's global bound on $f^{\prime }$, use the bounded-derivative Burgers flux $f(z)=z^2/2$ on $|z|\le 1$ and $f(z)=|z|-1/2$ outside. It is $C^1$ and $|f^{\prime }|\le 1$. Take data $-1$ for $x<0$, $1$ for $x>0$. A stationary jump obeys the condition because $f(-1)=f(1)$. Another weak solution is the rarefaction waveThe fan solves the equation inside each smooth region, is continuous at its edges, and has the same local $L^1$ initial trace. Thus weak solutions are not unique. The stationary expansion shock violates the entropy inequality, whereas the fan is admissible; using the unmodified quadratic flux alone would miss the printed global bounded-derivative hypothesis.

Let $L=\Delta +c$ with real $c$, and letThe Fredholm alternative for an elliptic Dirichlet problem says that $N$ is finite-dimensional. If $N=0$, all the prescribed data admit a unique solution. In general the exact boundary compatibility in the self-adjoint Fredholm alternative isThese conditions are necessary and sufficient; when they hold, all solutions form $u_0+N$. Thus in the second branch some data are incompatible, while compatible data have nonunique solutions. The outward normal derivative fixes the displayed sign.

To prove the alternative, put $X=\{w\in C^{2,\alpha }(\overline{\Omega }):w{|}_{\partial \Omega }=0\}$ and $Y=C^{0,\alpha }(\overline{\Omega })$. The standard zero-boundary Poisson equation theorem makes $\Delta :X\to Y$ an isomorphism. Multiplication by $c$, followed by ${\Delta }^{-1}$, is a compact operator $K:X\to X$: the embedding $C^{2,\alpha }\hookrightarrow C^{0,\alpha }$ is compact, and multiplication is bounded. Writing $u=\psi +w$ givesThe Fredholm alternative for a compact operator implies finite kernel and cokernel of equal dimension, and invertibility exactly when the kernel vanishes.

For the explicit compatibility conditions, use the self-adjoint Dirichlet realization of $L$ on $L^2$. It has compact resolvent: a sufficiently large positive shift of $-L$ is coercive, its inverse gains two derivatives, and the Rellich-Kondrachov compactness theorem makes that inverse compact. The Fredholm solvability condition for a self-adjoint operator is $F\perp N$. Standard boundary elliptic regularity upgrades the resulting weak solution to $C^{2,\alpha }$ for these data and boundary. Thus it gives the same kernel and solvability as the preceding Hölder-space formulation.

Finally Green second identity, with $z=0$ on the boundary and $Lz=0$, givesConsequently $\int zF=\int zf+\int \psi {\partial }_{\nu }z$, proving both necessity and sufficiency with the stated sign. If using complex data and real $c$, replace $z$ by $\overline{z}$ in the pairings.

Work first with essential bounds. On a Euclidean ball $B_R(x_0)⋐B_1$, put $M={\mathrm{esssup}}_{B_R}u$, $m={\mathrm{essinf}}_{B_R}u$ and $\omega =M-m$. The functions $M-u$ and $u-m$ are nonnegative weak solutions, hence weak supersolutions. In $B_{R/2}$, at least one of the sets $\{u\le (M+m)/2\}$ and $\{u\ge (M+m)/2\}$ has at least half the measure.

In the first case, the Weak Harnack inequality applied to $M-u$ givesIn the second case, applying it to $u-m$ gives the same lower bound for the improvement of the minimum. Thus the oscillation decay estimate isThis measure argument works even when $q<1$, avoiding a false use of the triangle inequality in that range.

Iteration yields ${\mathrm{osc}}_{B_r(x_0)}u\le C(r/R{)}^{\mu }{\mathrm{osc}}_{B_R(x_0)}u$, with, for example, $\mu =\min \{1/2,-\log \tau /\log 2\}\in (0,1)$. At Lebesgue points the shrinking oscillations define a unique continuous representative, and the estimate extends to every point by continuity. For $x,y\in B_{1/4}$ at distance less than $1/4$, apply the estimate to a Euclidean ball centred at $x$ with initial radius $1/2$ and final radius comparable to $|x-y|$. This Euclidean ball stays in $B_1$. Larger distances use the trivial bound $2\Vert u{\Vert }_{\infty }$. We obtainThe constants depend only on $n,\lambda ,\Lambda$. This derives the relevant conclusion of the De Giorgi-Nash-Moser theorem directly from part (a).

Specify an initial value or initial probability distribution. A strong solution of a stochastic differential equation uses a prescribed Brownian motion on a prescribed filtered probability space. It is continuous, adapted to the completed filtration generated by that Brownian motion and the initial value, and satisfiesalmost surely for every $t$, with ${\int }_0^t(|b(X_s)|+\sigma (X_s{)}^2)ds<\infty$ almost surely. The given initial variable is independent of future Brownian motion increments. Some courses allow a larger prescribed filtration in their definition of a strong solution of a stochastic differential equation; the Brownian-generated convention states explicitly what absence of extra randomness means.

A weak solution of a stochastic differential equation consists of a filtered probability space, a Brownian motion relative to its filtration, and a continuous adapted process $X$ satisfying the same integrability and integral equation. Here the space and noise are part of the unknown; the filtration may contain randomness beyond the initial variable and the driving path.

Pathwise uniqueness means that on any common filtered probability space, any two solutions with the same driving Brownian motion and the same initial value are indistinguishable. Uniqueness in law means that all weak solutions with the same initial probability distribution have the same law as path-valued random variables, even on different spaces. Pathwise uniqueness compares shared-noise paths; uniqueness in law compares path distributions. Neither uniqueness notion by itself asserts existence.

The Tanaka equation is an example:It has weak solutions and uniqueness in law, but lacks pathwise uniqueness. The value $s(0)=1$ is deliberate: using the sign function with value zero at zero would admit the identically zero solution and would change the example.

For a quick verification, take a Brownian motion $W$ and put $B=\int s(W)dW$. Its quadratic variation is $t$, so it is a Brownian motion by the Lévy characterization of Brownian motion, and associativity of stochastic integration gives $W=\int s(W)dB$. Every weak solution likewise has quadratic variation $t$ and therefore the law of Brownian motion. On this same enlarged filtration, both $W$ and $-W$ solve the equation driven by $B$: $s(-W)=-s(W)$ away from zero, and the Brownian zero set has zero time measure, so the discrepancy at zero contributes nothing to the Itô integral. These solutions differ with positive probability at any positive time. This proves the claimed distinction without relying on a choice of zero convention left unstated.

Use the real Hilbert space $V=H_0^1(0,1)$, the zero-boundary Sobolev space, with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincaré inequality makes this equivalent to its usual Sobolev space norm. Define the symmetric bilinear form and continuous linear functionalThe Cauchy-Schwarz inequality and the sharp interval Poincaré inequality implyThus $a$ is a bounded bilinear form and a coercive bilinear form, and $\ell$ is bounded because $f$ is square-integrable. The Lax-Milgram theorem supplies a unique weak solution:This is obtained from the differential equation by integration by parts; the test functions have zero boundary trace. Conversely, the weak solution satisfies $y^{\prime \prime }=xy-f$ in the sense of distributions. Since $xy-f\in L^2$, it belongs to $H^2(0,1)\cap H_0^1(0,1)$ and solves the equation almost everywhere.

Equivalently, the Ritz method minimizesover $V$. Indeed $J(y+w)-J(y)=a(w,w)/2$, since the mixed term is $a(y,w)-\ell (w)=0$. This proves existence and uniqueness of the minimizer directly from the weak solution, as well as its equivalence to the variational problem.

Take the difference of two weak solutions, so $f={\psi }_0={\psi }_1=0$. At the stated $H^1(U_T)$ regularity, it is not justified simply to test by $u_t$, which need not belong to the spatial test space $H_0^1$. Instead use the antiderivative test underlying uniqueness of Sobolev weak solutions of a wave equation.

Fix $s\in (0,T]$ and defineThese are Bochner integrals in $H_0^1$. The function $v$ is an admissible space-time $H^1$ test, $v_t=-u$ before $s$, and $v(t)=z(s)-z(t)$. The time term in the weak formulation isbecause $u(0)=0$. Symmetry of $a^{ij}$ and integration by parts in time giveFor the lower-order part, integration by parts in space givesBounded coefficients, the Poincaré inequality for $v$, and the Young inequality now implySince $Dv(t)=Dz(s)-Dz(t)$,For $s$ in a sufficiently short fixed interval, absorb the last term into the left-hand side. The Gronwall inequality applied to $\Vert u(s){\Vert }_2^2+(\theta /2)\Vert Dz(s){\Vert }_2^2$ proves $u=0$ on that interval. The equation gives continuity of $u_t$ in $H^{-1}$, so both initial traces at its endpoint are again zero. Repeating with the same uniform coefficient bounds covers $[0,T]$. Therefore

Work in the real zero-boundary Sobolev space $X=H_0^1(0,1)$, whose members have square-integrable weak derivatives and zero endpoint trace. Define the bounded bilinear form and linear functionalThe weak formulation is to find $u\in X$ with $a(u,v)=\ell (v)$ for every $v\in X$. Integration by parts recovers the differential equation for a smooth solution; conversely, this identity is the definition of its weak solution.

Use the full $H^1$ norm. The Cauchy-Schwarz inequality gives $|a(u,v)|\le \Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$ and $|\ell (v)|\le \Vert v{\Vert }_{H^1}/\sqrt{3}$. Also $a(v,v)=\Vert v{\Vert }_{H^1}^2$, so $a$ is a coercive bilinear form with constant one. The Lax-Milgram theorem states that a bounded, coercive bilinear form on a real Hilbert space determines a unique weak solution for every bounded linear functional. It therefore applies here.

The equivalent minimization problem isIf $u$ is the weak solution, symmetry and $a(u,v-u)=\ell (v-u)$ giveThis is positive unless $v=u$, proving a unique global minimizer. Conversely, differentiating $\mathcal{E}(u+tv)$ at $t=0$ gives the weak identity. Thus the unique energy minimizer is exactly the unique weak solution. As a check, the classical representative isIt satisfies both endpoint conditions and $-u^{\prime \prime }+u=x$.

The Dirichlet boundary condition is built into the first component's space, while the Neumann boundary condition is natural. Thus a weak solution is a pairsuch that for every $(v,z)\in H_0^1(U)\times H^1(U)$,If $u,w$ are $C^2$ up to the boundary, taking compactly supported test functions and applying the fundamental lemma of the calculus of variations gives both differential equations pointwise in $U$. Membership of $H_0^1(U)$ gives $u=0$ on $\partial U$. Applying integration by parts to the second identity and using its differential equation leavesfor every smooth boundary trace $z$. Hence $\partial w/\partial \nu =0$ on $\partial U$, so the equations and both boundary conditions hold classically.

Consider a bounded Lipschitz domain $\Omega$ and homogeneous Dirichlet boundary conditions forAssume $A$ is a bounded real symmetric matrix field with ${\xi }^{T}A(x)\xi \ge a_0|\xi {|}^2$ uniformly, $c\ge 0$ is bounded, and $f$ defines a bounded linear functional on $V=H_0^1(\Omega )$, the zero-boundary Sobolev space. The weak formulation isThe Cauchy-Schwarz inequality gives continuity $|a(u,v)|\le M\Vert u{\Vert }_V\Vert v{\Vert }_V$. Uniform ellipticity and the Poincaré inequality give a coercive bilinear form, $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$. The Lax-Milgram theorem therefore supplies a unique weak solution, with $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\prime }}/\alpha$. Its proof represents $a$ by a bounded linear operator $T$ using the Riesz representation theorem; coercivity gives an injective operator with closed image, and a zero orthogonal complement makes that image all of $V$. This yields existence, uniqueness and the displayed estimate.

For a conforming finite element space $V_h\subseteq V$, the Galerkin method finds $u_h\in V_h$ such that $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. With a basis ${\varphi }_1,\dots ,{\varphi }_N$, this is the linear system $KU=F$, where the stiffness matrix is $K_{ij}=a({\varphi }_j,{\varphi }_i)$ and $F_i=\ell ({\varphi }_i)$. Its positive-definite matrix property proves unique discrete solvability. Local support gives a sparse matrix through elementwise assembly.

For the symmetric problem, the Ritz method minimizes $J(v)=a(v,v)/2-\ell (v)$ over $V_h$. Differentiating in every direction $v_h$ gives precisely the Galerkin method equations. Conversely, if $u_h$ solves them, $J(u_h+w_h)-J(u_h)=a(w_h,w_h)/2$, proving it is the unique minimum. The same argument over $V$ gives $J(v)-J(u)=\Vert v-u{\Vert }_a^2/2$, where $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ is the energy norm. Thus the Ritz method chooses the best trial function in that norm.

Subtracting the continuous and discrete equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For arbitrary $v_h\in V_h$ and $e=u-u_h$, it follows that $a(e,e)=a(e,u-v_h)$. Coercivity and continuity prove the Céa lemma:In the symmetric energy norm, the Cauchy-Schwarz inequality improves the constant to one. Equivalently, Galerkin orthogonality gives the Pythagorean identity $\Vert u-v_h{\Vert }_a^2=\Vert u-u_h{\Vert }_a^2+\Vert u_h-v_h{\Vert }_a^2$. These estimates establish stability and show that approximation properties of the trial space determine convergence.

For continuous piecewise-linear elements on a shape-regular mesh in dimensions at most three, a finite element interpolation estimate gives $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ when $u\in H^2(\Omega )$. Combining it with the Céa lemma yields $O(h)$ energy norm error. More generally, degree-$p$ elements give $O(h^p)$ error in $H^1$ under the corresponding regularity and approximation assumptions. Mere mesh refinement cannot supply an order whose required solution regularity is absent.

An additional order in the L2 norm follows under a dual elliptic regularity assumption. Solve $a(v,z)=(e,v{)}_{L^2}$ and assume $\Vert z{\Vert }_{H^2}\le C\Vert e{\Vert }_{L^2}$. The Aubin–Nitsche duality argument uses Galerkin orthogonality to obtainTherefore $\Vert e{\Vert }_{L^2}\le Ch\Vert e{\Vert }_{H^1}$, giving $O(h^2)$ for the piecewise-linear case with $u\in H^2$. This improvement explicitly requires regularity of the dual problem.

As a concrete example, solve $-u^{\prime \prime }=1$ on $(0,1)$ with zero endpoint values. On a uniform mesh of spacing $h$, the piecewise-linear hat functions give $K_{jj}=2/h$, $K_{j,j\pm 1}=-1/h$ and $F_j=h$. The discrete equations are $2U_j-U_{j-1}-U_{j+1}=h^2$, whose solution is $U_j=x_j(1-x_j)/2$. The exact solution is $u(x)=x(1-x)/2$, so $u_h$ is its piecewise-linear nodal interpolant. On each interval $[x_j,x_{j+1}]$,Integrating the squared error and squared derivative error gives the explicit rateswhere the $H_0^1$ norm here is $\Vert v^{\prime }{\Vert }_{L^2}$.

For nonsymmetric coercive problems, the Galerkin method, Galerkin orthogonality and the Céa lemma still apply, but the symmetric quadratic minimization interpretation of the Ritz method generally does not. Nonconforming trial spaces, inexact integration and noncoercive equations need additional arguments beyond the conforming coercive theory proved here.

Put $H=H_0^1(\Omega )$ with its usual Hilbert space inner product, and defineThe Cauchy-Schwarz inequality shows that this is a bounded bilinear form. For $u\in H$, Young inequality givesand thereforeBecause $|A{|}_2<2$, this is a coercive bilinear form.

The map $u\mapsto {\int }_{\Omega }uf$ is a bounded linear functional on $H$. By the Riesz representation theorem, it equals $⟨u,F{⟩}_H$ for some $F\in H$. Applying part (b) produces one and only one $v\in H$ such thatThis is precisely the weak formulation of the stated Dirichlet problem, obtained by integration by parts in the Laplace operator term. Thus $v$ is its unique weak solution; this is the constant-drift massive-Laplacian Dirichlet problem.

Use the Poisson integral to choose a smooth harmonic function $G$ on the unit disk whose boundary value is $g$. Seek $v$ in the form $v=w+G$, which is the standard homogenization of Dirichlet boundary data. Then $w\in H_0^1(\Omega )$ and, since $\Delta G=0$,Part (d) gives a unique such $w$, and henceis the unique weak solution with boundary value $g$.

Choose the Sobolev space $V$ encoding the homogeneous essential boundary conditions, setafter the appropriate integration by parts, and define the energy functionalIts first variation is $J^{\prime }(u)v=a(u,v)-\ell (v)$, so its stationary points are exactly the solutions of the weak formulationIf $a$ is bounded, symmetric, and coercive and $\ell \in V^{\prime }$, the Lax-Milgram theorem supplies a unique weak solution $u$. Moreover, for every $w\in V$,unless $w=0$. Thus $J$ is strictly convex, and $u$ is its unique global minimizer. This proves existence and uniqueness of the minimizer and of the weak solution simultaneously.

The zero-boundary Sobolev space isOn it defineIf this quadratic form vanishes, then $\nabla u=0$. The Poincaré inequality givesso $u=0$. It is therefore an inner product, and its norm is equivalent to the usual $H^1$ norm.

A function $u\in H_0^1(\Omega )$ is a weak solution whenThis follows from integration by parts and incorporates the homogeneous Dirichlet boundary condition through membership in $H_0^1$.

By the contraction mapping theorem, the contraction $A:X_{b,\tau }\to X_{b,\tau }$ has a unique fixed point $u$. The identity $Au=u$ says exactly that $u$ is the weak solution of the linear initial boundary value problem whose forcing is $u^2\sin u$. Consequently $u\in X_{\tau }$ satisfieswith the prescribed initial and homogeneous Dirichlet data. Thus a sufficiently small $\tau >0$ gives a local weak solution of the semilinear wave equation, as summarized by the local weak solution by contraction for a semilinear wave equation.